QUESTION 7 A clothing store produces two models of dresses: a style for adults (
ID: 339820 • Letter: Q
Question
QUESTION 7 A clothing store produces two models of dresses: a style for adults (A) and a style for Kids (K). Each style uses the same fabric, of which there is 1,000 yards for production each week, and both require assembly time, of which there is 60 hours per week available. Each adult dress requires 3 yards of fabric and it takes 10 minutes to assemble. For each kid's dress, it requires 1.5 yards of fabric and it takes 6 minutes to assemble. Which of the following is not a feasible solution when formulating this as a linear program to maximize profit? b.A0; K- 600 Oc. A 300; K 100 OdA 200; K 200 Oe, A = 330; K = 0Explanation / Answer
let A & K be the two models of dress
Constraint
Fabric (Yard )
3A + 1.5K =< 1000 -----1
10A + 6K =< 3600 ----2
Feasible solution are the ones which should staisfy the two above constraint
a) A = 0, K= 0
Putting in eq 1 and 2
Eq 1 , 3A + 1.5K =< 1000
0 =< 1000 Which is true
Eq 2 , 10A + 6K =< 3600
10A + 6K =< 3600
0 =< 3600 Which is true
As both constraint is staisfied by point so , it is feasible
b)
A= 0, K = 600
Putting in eq 1 and 2
Eq 1 , 3A + 1.5K =< 1000
900 =< 1000 ---- True
Eq 2 , 10A + 6K =< 3600
3600 =< 36000 ---- True
As it satisfy both constraint So feasible solution
c) A= 300, K =100
Eq1 , 3A + 1.5K =< 1000
1050 =< 1000 --- False
Eq 2 10A + 6K =< 3600
3600 =< 3600 ---- True
as it does not satisfy the constraint of Fabric so it is not feasible
D) A= 200, K =200
Eq1 3A + 1.5K =< 1000
900 =< 1000 ---True
Eq 2 10A + 6K =< 3600
3200 =< 3600 ---- True
As it satisfy both constraint So feasible solution
e) A = 330, K = 0
Eq1 , 3A + 1.5K =< 1000
990 =< 1000 -- True
Eq 2 10A + 6K =< 3600
3300 =< 3600 -- True
As it satisfy both constraint so it is feasbile solution.
Answer : C) A = 300, K =100 is not feasible solution
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